Cyclic Array Maximization — Problem Statement & Solution Guide
Problem Description
You are given a circular array of integers resources and an integer missions, representing the limit of space missions. Find the maximum sum of resources that can be collected by traversing the array in a cyclic manner.
Examples
Input
[1, 2, 3, 4, 5], 3
Output
15
Explanation: Step-by-step: We can collect 15 resources by traversing the array in a cyclic manner starting from index 0, then 1, then 2, and finally 0 again. This is because the prefix sum array is [1, 3, 6, 10, 15, 15] and the maximum sum of resources that can be collected is 15.
Input
[5, 4, 3, 2, 1], 3
Output
15
Explanation: Step-by-step: We can collect 15 resources by traversing the array in a cyclic manner starting from index 0, then 1, then 2, and finally 0 again. This is because the prefix sum array is [5, 9, 12, 14, 15, 15] and the maximum sum of resources that can be collected is 15.
Constraints
- {"name":"resources length","type":"integer","min":1,"max":1000}
- {"name":"missions","type":"integer","min":1,"max":7}
- {"name":"resource values","type":"integer","min":-1000,"max":1000}
Optimal Approach & Strategy
The optimized approach uses prefix sums and cycle detection to find the maximum resources that can be collected in O(n * missions) time complexity by considering all possible start and end points.
Brute Force Approach
The brute-force approach involves trying all possible combinations of missions and calculating the total resources collected for each combination, resulting in a time complexity of O(n^2 * missions).
Verified Code Solutions
public int cyclicArrayMaximization(int[] resources, int missions) {
if (resources.length == 0 || missions == 0) {
return 0;
}
if (resources.length == 1) {
return resources[0] * Math.min(missions, 1);
}
int[] prefixSum = new int[resources.length + 1];
for (int i = 0; i < resources.length; i++) {
prefixSum[i + 1] = prefixSum[i] + resources[i];
}
int maxSum = 0;
for (int i = 0; i < resources.length; i++) {
maxSum = Math.max(maxSum, prefixSum[i + missions] - prefixSum[i]);
}
return maxSum;
}def cyclic_array_maximization(resources, missions):
if len(resources) == 0 or missions == 0:
return 0
if len(resources) == 1:
return resources[0] * min(missions, 1)
prefix_sum = [0] * (len(resources) + 1)
for i in range(len(resources)):
prefix_sum[i + 1] = prefix_sum[i] + resources[i]
max_sum = 0
for i in range(len(resources)):
max_sum = max(max_sum, prefix_sum[i + missions] - prefix_sum[i])
return max_sumAsked in Top Tech Interviews
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